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(A) The fundamental frequency of a closed pipe $A$ of length $L_1$ is given by $n_1 = \frac{v}{4L_1}$.For an open pipe $B$ of length $L_2$,the frequen

Vedclass · Accepted Answer

$1: 6$

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(A) The fundamental frequency of a closed pipe $A$ of length $L_1$ is given by $n_1 = \frac{v}{4L_1}$.For an open pipe $B$ of length $L_2$,the frequencies are $n = \frac{mv}{2L_2}$ where $m = 1, 2, 3, \dots$. The first overtone is $m=2$ and the second overtone is $m=3$.Thus,the frequency of the second overtone of pipe $B$ is $n_2 = \frac{3v}{2L_2}$.Given that the frequencies are in unison,$n_1 = n_2$.Therefore,$\frac{v}{4L_1} = \frac{3v}{2L_2}$.Rearranging the terms to find the ratio $\frac{L_1}{L_2}$,we get $\frac{L_1}{L_2} = \frac{2}{4 	imes 3} = \frac{2}{12} = \frac{1}{6}$.So,the ratio is $1:6$.

The fundamental frequency of an air column in pipe $A$ closed at one end is in unison with the second overtone of an air column in pipe $B$ open at both ends. The ratio of the length of the air column in pipe $A$ to that of the air column in pipe $B$ is:

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